1. Decay Problem

"The half-life of radioactive carbon (14) is 5730 years. If the amount of radioactive carbon left over after 1000 years is 5 grams, how much radioactive carbon did you start with?"

equation given to me is $\displaystyle A = A_0e^{-kt}$

Now how do I find $\displaystyle A_0$

I've found $\displaystyle k = -0.000121$

Therefore I have formulated this equation

$\displaystyle 5 = A_0e^{(-0.000121*1000)}$

My problem is, how do I find $\displaystyle A_0$ from the above formula or have I done something wrong. Everytime I do it, I end up with weird answer. Please help

Just out of curiosity though, my answer = 5.643124562. Is that right? or wrong?

2. you know how to solve that, as you did since you got a correct answer

3. Originally Posted by dwat
"The half-life of radioactive carbon (14) is 5730 years. If the amount of radioactive carbon left over after 1000 years is 5 grams, how much radioactive carbon did you start with?"

equation given to me is $\displaystyle A = A_0e^{-kt}$

Now how do I find $\displaystyle A_0$

I've found $\displaystyle k = -0.000121$ Mr F says: There should be no negative. Clearly a typo on your part. And you should state the accuracy that you've rounded your answer to.

Therefore I have formulated this equation

$\displaystyle 5 = A_0e^{(-0.000121*1000)}$

My problem is, how do I find $\displaystyle A_0$ from the above formula or have I done something wrong. Everytime I do it, I end up with weird answer. Please help

Just out of curiosity though, my answer = 5.643124562. Is that right? or wrong?
$\displaystyle 5 = A_0 e^{-(0.000121)(1000)} = A_0 e^{-0.121} \Rightarrow A_0 = \frac{5}{e^{-0.121}} = 5.64$ gram, correct to three significant figures.

Since you have rounded your calculated value of k to three significant figures, your final answer is calculator madness - you cannot have an answer that is more accurate than the least accurate data used to get the answer ....

4. Originally Posted by mr fantastic
$\displaystyle 5 = A_0 e^{-(0.000121)(1000)} = A_0 e^{-0.121} \Rightarrow A_0 = \frac{5}{e^{-0.121}} = 5.64$ gram, correct to three significant figures.

Since you have rounded your calculated value of k to three significant figures, your final answer is calculator madness - you cannot have an answer that is more accurate than the least accurate data used to get the answer ....
So I did get it right

Oh sorry, yeah I meant k = 0.000121

Well it asked to round k to 6 decimal places so I did. But you have a point. My mistake